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Quick look over friction pulley system

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Objects with masses m1 = 12.0 kg and m2 = 4.00 kg are connected by a light string that passes over a frictionless pulley. If, when the system starts from rest, m2 falls 1.00 m in 1.70 s, determine the coefficient of kinetic friction between m1 and the table.



    2. Relevant equations

    Fg = Fn = mg
    F = ma
    μ = Ff/Fn

    3. The attempt at a solution

    Fg = mg
    Fg = 16*9.8
    Fg = 156.8 = Fn

    F = ma [m2]
    = 4*9.8
    = 39.2

    ƩFx = m2 - Ff (sum of all forces in the x direction is = m2 - Frictional force)
    = 39.2 - Ff
    Ff = 39.2

    μ = Ff/Fn
    μ = 39.2/156.8
    μ = .25

    I tried entering the coefficient of kinetic friction into the "submit box" but the answer is incorrect.
    could somebody please point out my mistake?
     
  2. jcsd
  3. Oct 23, 2012 #2

    ehild

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    Is there a table in the problem??? A drawing would be nice, with free body diagram for both masses.

    ehild
     
  4. Oct 23, 2012 #3
    this is the picture i have
     

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  5. Oct 23, 2012 #4

    ehild

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    Nice picture:smile:.
    Both blocks accelerate with the same amount as they are connected with a spring.
    m2 falls 1 m in 1.7 s. What is its acceleration?

    ehild
     
  6. Oct 23, 2012 #5
    thanks :)
    and acceleration is:
    Δx = .5 a t^2
    1 = .5(a)1.7^2
    1 = 1.445a
    .69 = a
     
  7. Oct 23, 2012 #6
    Your mistake is in your calculation of normal force. Only m1 should factor into normal force. M2 experiences only gravity and tension :)
     
  8. Oct 23, 2012 #7
    i multiply m1 by .69?
     
  9. Oct 23, 2012 #8

    ehild

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    It is OK,
    but do not round off during the calculations, keep one more digit as the number of the significant digits in the data. And show the unit. So a=0.6920 m/s2

    The acceleration of m2 is know, so the net force on m2 is F=m2a. What forces act on m2?

    ehild
     
  10. Oct 23, 2012 #9
    gravity and tension?
     
  11. Oct 23, 2012 #10

    ehild

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    If you multiply m1 with the common acceleration, you get the net horizontal force on m1. That force is the resultant of what forces?

    Think of the string. Does it exert force?

    ehild
     
  12. Oct 23, 2012 #11
    the string which is connected to m2 pulls m1. and then there is friction on m1
     
  13. Oct 23, 2012 #12

    ehild

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    Yes. Gravity and tension of the rope result in F=m2a. Can you find the magnitude of the tension?

    ehild
     
  14. Oct 23, 2012 #13

    ehild

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    Yes, what are the direction of these forces?

    ehild
     
  15. Oct 23, 2012 #14
    gravity on m2 = 39.2 N. the tension = 39.2 also?
    and friction is in the "negative" direction
    force from m2 is in the "positive" direction
     
  16. Oct 23, 2012 #15

    ehild

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    The tension in the rope acts upward on m2. See the picture, I added the forces. So the net force on m2 is m2g-T=m2a. What is T then?

    ehild
     

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  17. Oct 23, 2012 #16
    36.432 = t
     
  18. Oct 23, 2012 #17

    ehild

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    Yes, the tension is T=36.432 N.
    Now what forces act on m1?

    ehild
     
  19. Oct 23, 2012 #18
    Tension and friction?
     
  20. Oct 23, 2012 #19

    ehild

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    Yes. The tension is the same at both ends of the string. You know acceleration, mass and one force, tension. How much is the other force, the force of friction?

    ehild
     
  21. Oct 23, 2012 #20
    ƩFx = T - Ff
    ƩFx = 36.432 - Ff
    Ff = 36.432 N
     
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